Problem: Is ${726513}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {726513}= &&{7}\cdot100000+ \\&&{2}\cdot10000+ \\&&{6}\cdot1000+ \\&&{5}\cdot100+ \\&&{1}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {726513}= &&{7}(99999+1)+ \\&&{2}(9999+1)+ \\&&{6}(999+1)+ \\&&{5}(99+1)+ \\&&{1}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {726513}= &&\gray{7\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {7}+{2}+{6}+{5}+{1}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${726513}$ is divisible by $9$ if ${ 7}+{2}+{6}+{5}+{1}+{3}$ is divisible by $9$ Add the digits of ${726513}$ $ {7}+{2}+{6}+{5}+{1}+{3} = {24} $ If ${24}$ is divisible by $9$ , then ${726513}$ must also be divisible by $9$ ${24}$ is not divisible by $9$, therefore ${726513}$ must not be divisible by $9$.